如何在jquery

Question

所以我已经在这里问了一些问题并解决了我的问题，但我认为它们正在增长而不是下降，哈哈。我现在的问题是我得到了＆＃34; Undefined index＆＃34; error.I会留下错误here的图片。我试图添加变量来定义我的fsearch.Tried之类的事情

$fsearch = mysqli_escape_string($connect, $_POST['fsearch']);

更简单就像。

$fsearch = $_POST['fsearch'];

但它不起作用。这是代码。

＆＃13;

＆＃13;

jQuery(document).ready(function ($) {
	$("#food_search").keyup(function(event){
		var search_term =$(this).val();
$.ajax({
	type:"POST",
	url:"/Food-Search",
	data:{fsearch:search_term},
	success:function(res){
		$("#food_search_result").html(res);
	},
	error: function (xhr, ajaxOptions, thrownError) {
           alert(xhr.status);
           alert(xhr.responseText);
           alert(thrownError);
       }
        });
	});
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<!----------------------------------------------------------------
                              HTML
----------------------------------------------------------------->

<form method="POST">
<p>Търсене на храни: <input type="text" name="fsearch" id="food_search"></p>
</form>
<!----------------------------------------------------------------
                             PHP
-----------------------------------------------------------------><?php

$hostname = "localhost";
$username = "name";
$password = "password";
$databaseName = "dbName";
$connect = mysqli_connect($hostname, $username, $password, $databaseName);

if (!empty($_POST['fsearch'])) {
   $fsearch = mysqli_escape_string($connect, $_POST['fsearch']);
   $req = $connect->prepare('SELECT * FROM food_data_bg WHERE title LIKE "%".$fsearch."%"');
   $req->execute(array(
     'fsearch'=>'%'.$_POST['fsearch'].'%'
));
   if($req->rowCount()==0){
    echo 'Не бяха намерени резултати!';
}
else{
while($foodsearch=$req->fetch()){
?>
<div class="search-result">
    <img src="<?php echo $data['fimage']; ?>" class="fimage"/>
    <span class="result-title"><?php echo $data['title'];?></span><br>
    <span class="calories-total"><?php echo $data['calories total'];?></span><br>
</div>
<?php
}
}
}
?>

＆＃13;

＆＃13;

＆＃13;

代码用于搜索，它必须在每个keyup上获取#food-search（输入字段）的值并将其发送给PHP，PHP必须检查数据库中的匹配并打印它。（我想我得到了通过在PHP页面中打印结果并将其返回到java以在#food_search_result div）

中打印它还有1个错误

Answer 1

希望这会有所帮助。

PHP代码

$hostname = "localhost";
$username = "root";
$password = "";
$databaseName = "stackoverflow";
if (!empty($_POST['fsearch'])) {    
$fsearch = $_POST['fsearch'];
$mysqli = new mysqli($hostname , $username, $password, $databaseName); 
if($mysqli->connect_error)
{
die("$mysqli->connect_errno: $mysqli->connect_error");
}
$query = "SELECT * FROM food_data_bg where title LIKE ?";
$stmt = $mysqli->stmt_init();
if(!$stmt->prepare($query))
{
print "Failed to prepare statement\n";
}
else
{
$searchVar = "%" . $fsearch . "%";
$stmt->bind_param("s", $searchVar);

$stmt->execute();
$result = $stmt->get_result();

if(!$result->num_rows){
echo 'Не бяха намерени резултати!';
}       
while ($row = $result->fetch_array())
{            
?>
<div class="search-result" style='clear:both; border:1px solid red;'>
<img src="<?php //echo $row['fimage']; ?>" class="fimage"/>
<span class="result-title"><?php echo $row['title'];?></span><br>
<span class="calories-total"><?php echo $row['calories total'];?></span><br>
</div>
<?php        
}     
}
$stmt->close();
$mysqli->close();
}

HTML代码

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
jQuery(document).ready(function ($) { 
    $("#food_search").keyup(function(event){
        var search_term =$(this).val();
        console.log(search_term);
$.ajax({
    type:"POST",
    url:"Food-Search.php",
    data:{fsearch:search_term},
    success:function(res){
        $("#food_search_result").html(res);
        console.log(res);
    },
    error: function (xhr, ajaxOptions, thrownError) {
           alert(xhr.status);
           alert(xhr.responseText);
           alert(thrownError);
       }
        });
    });
});
</script>

<!----------------------------------------------------------------
                              HTML
----------------------------------------------------------------->



<form method="POST">
<p>Търсене на храни: <input type="text" name="fsearch" id="food_search"></p>
</form>
<!----------------------------------------------------------------
                             PHP
----------------------------------------------------------------->

<div style="border:#FF0000;clear:both; padding:10px;min-height:300px;" id="food_search_result"></div>